Answer:
θ ≈ 65°
Explanation:
From the question given above, the following data were obtained:
Maximum height (H) = 8 m
Range (R) = 15 m
Initial velocity (u) = v
Angle θ =?
H = u²Sine²θ / 2g
8 = v²Sine²θ / 2g
Cross multiply
8 × 2g = v²Sine²θ
16g = v²Sine²θ
Divide both side by Sine²θ
v² = 16g / Sine²θ.... (1)
R = u²Sine2θ / g
15 = v²Sine2θ / g
Cross multiply
15 × g = v²Sine2θ
15g = v²Sine2θ
Divide both side by Sine2θ
v² = 15g / Sine 2θ... (2)
Summary:
v² = 16g / Sine²θ.... (1)
v² = 15g / Sine 2θ... (2)
Equate equation 1 and 2
16g / Sine²θ = 15g / Sine 2θ
16 / Sine²θ = 15 / Sine 2θ
Recall:
Sine²θ = SineθSineθ
Sine 2θ = 2SineθCosθ
16 / Sine²θ = 15 / Sine 2θ
16 / SineθSineθ = 15 / 2SineθCosθ
16 / Sineθ = 15 / 2Cosθ
Cross multiply
15 × Sineθ = 16 × 2Cosθ
15Sineθ = 32Cosθ
Divide both side by Cosθ
15Sineθ / Cosθ = 32
Divide both side by 15
Sineθ / Cosθ = 32/15
Recall:
Sineθ / Cosθ = Tanθ
Sineθ / Cosθ = 32/15
Tanθ = 32/15
Tanθ = 2.1333
Take the inverse of Tan
θ = Tan¯¹ 2.1333
θ ≈ 65°
Answer:
a) τ₁ = 660 N m, b) τ’= 686 N m, c) F = 623.6 N
Explanation:
a) For this exercise let's use the concepts of torque and rotational balance.
For this we set a reference system at the base and assuming that the counterclockwise rotations are positive
where the force F = 600 N, the distance to the axis is x = 1.1 m, the mass of the system m = 70g and the weight is placed at the point of the center of gravity x_{cm} = -1.0 m
The torque at the front is
τ₁ = F x
τ₁ = 600 1.1
τ₁ = 660 N m
b) let's write the rotational equilibrium condition
∑ τ = 0
τ'- W x_{cm} = 0
τ ’= mg x_{cm}
τ’= 70 9.8 1.0
τ’= 686 N m
c) the greatest force Matt can apply
τ’= F x
F = τ’/ x
F = 686 / 1.1
F = 623.6 N
Answer:
α = 11.05 rad/s^{2}
Explanation:
torque (T) = 2.1 N.m
moment of inertia (I) = 0.19 kg.m^{2}
angular acceleration (α) = ?
from torque = moment of inertia x angular acceleration
2.1 = 0.19 x α
α = 11.05 rad/s^{2}
Answer:
2.398
Explanation:
In order to solve this problem we are required to use the following information and the question expects us to give depth at which theask is seen.
Dn = D x nobservation/n object
This is the formula for apparent depth
Depth = 3.19
The index of refraction for water is 1.33
3.19x1/1.33
= 3.19/1.33
= 2.398m
the person see his dive mask at
2.398m (apparent depth in meters
Answer:
1. What conclusions can you draw about the relationship between the temperature of a ball and its bounce height? Is there a direct or inverse relationship? Write an evidence-based claim.